The differentiable functions $x$ and $y$ are related by the following equation: $\dfrac{1}{y}=\cos(x)$ Also, $\dfrac{dx}{dt}=-2$. Find $\dfrac{dy}{dt}$ when $x=\pi$.
Explanation: Let's start by differentiating the equation $\dfrac{1}{y}=\cos(x)$ with respect to $t$. $\begin{aligned} \dfrac{1}{y}&=\cos(x) \\\\ -\dfrac{1}{y^2}\cdot\dfrac{dy}{dt}&=-\sin(x)\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that ${\dfrac{dx}{dt}=-2}$, and we want to find $\dfrac{dy}{dt}$ when ${x=\pi}$. We also need to find the corresponding value of ${y}$. To do that, we can plug ${x=\pi}$ into the relating equation: $\begin{aligned} \dfrac{1}{{y}}&=\cos({x}) \\\\ \dfrac{1}{{y}}&=\cos({\pi}) \\\\ \dfrac{1}{{y}}&=-1 \\\\ {y}&{=-1} \end{aligned}$ Let's plug ${x=\pi}$, ${y=-1}$, and ${\dfrac{dx}{dt}=-2}$ into the equation we obtained: $\begin{aligned} -\dfrac{1}{{y}^2}\cdot\dfrac{dy}{dt}&=-\sin({x})\cdot{\dfrac{dx}{dt}} \\\\ -\dfrac{1}{({-1})^2}\cdot\dfrac{dy}{dt}&=-\sin({\pi})({-2}) \\\\ -\dfrac{dy}{dt}&=0 \\\\ \dfrac{dy}{dt}&=0 \end{aligned}$ In conclusion, when $x=\pi$, the value of $\dfrac{dy}{dt}$ is $0$.